# $\lt_{ip}$ is a well-defined well-ordering of iterable set premice.

I have cross-posted this question on MO. This is the link:https://mathoverflow.net/questions/351894/lt-ip-is-a-well-defined-well-ordering-of-iterable-set-premice

This question of mine arises from Kanamori's the higher infinite, where he tries to prove the result attributed to Silver and Solovay, that if $$\omega_1^{L[U]} = \omega_1$$, then there is a $$\Pi_2^1$$ set without the perfect set property.

Here we are dealing with ZFC$$^-$$(ZFC minus the Powerset Axiom) premice.

To be more precise, assume that $$M$$ is a transitive model of ZFC$$^-$$ and that $$U$$ is some set in $$V$$. We say that $$\langle M, \in, U\rangle$$ is a ZFC$$^-$$ premouse (at $$\kappa$$) iff $$U$$ is an $$M$$-ultrafilter over $$\kappa$$ and that for some $$\zeta$$, $$M = L_\zeta[U].$$

Also we say that two premice $$\langle M, \in, U\rangle$$ and $$\langle N, \in, W\rangle$$ are comparable iff $$\exists F \exists \zeta \exists \eta$$ such that $$M = L_\zeta[F]$$ and that $$N = L_\eta[F]$$.

Now there is a lemma (called the Comparison lemma) which states that:

If $$\langle M, \in, U\rangle$$ and $$\langle N, \in, W\rangle$$ are iterable premice, then they have iterates which are comparable.

Now let $$\langle M, \in, U\rangle$$ and $$\langle N, \in, W\rangle$$ be iterable premice, define $$\lt_{ip}$$ in the following manner:

$$\langle M, \in, U\rangle \lt_{ip}\langle N, \in, W\rangle$$ iff there exists some $$F$$ and some $$\zeta$$ and $$\eta$$ such that $$\langle L_\zeta[F], \in, F\cap L_\zeta[F]\rangle$$ is an iterate of $$\langle M, \in, U\rangle$$ and $$\langle L_\eta[F], \in, F\cap L_\eta[F]\rangle$$ is an iterate of $$\langle N, \in, W\rangle$$, such that $$\zeta \lt \eta$$.

Now Kanamori says that this ordering is a well-defined well-ordering of iterable set premice. And he says that this is straightforward to check, using the comparison lemma.

I am still stuck on showing that this is well-defined. The best idea I had was to show that: if $$\alpha$$ and $$\beta$$ are the first indices of the iterations of $$M$$ and $$N$$ which are comparable, then all the higher iterates should be comparable in a "coherent" fashion.

So what I did was this: Let $$\langle M_\alpha, \in, U_\alpha, \kappa_\alpha, i_{\alpha\beta} \rangle_{\alpha\le\beta\in\text{On}}$$ and $$\langle N_\alpha, \in, W_\alpha, \lambda_\alpha, j_{\alpha\beta} \rangle_{\alpha\le\beta\in\text{On}}$$ be the iterations of $$M$$ and $$N$$, respectively. Then let $$\alpha$$ and $$\beta$$ be the first ordinals where $$M_\alpha$$ and $$N_\beta$$ are comparable. Let $$F, \zeta, \eta$$ be such that: $$M_\alpha = L_\zeta[F]$$ and $$N_\beta = L_\eta[F]$$. There are $$2$$ cases:

$$(1)$$ $$\zeta = \eta$$: At this point I know that the rest of their iterations should be the same. But I think for totality's sake I have to show that $$M = N$$. Which is not obvious to me at the moment. (*)

$$(2)$$ $$\zeta \lt \eta$$: We can see that $$M_\alpha,U_\alpha \in N_\beta$$ so that the iteration of $$N$$ can witness the iteration of $$M$$ inside it. So for $$\delta \in \text{On}$$, we can see that $$M_{\alpha + \delta} \in N_{\beta + \delta}$$, but I can't generalize this argument for all $$\delta \ge \alpha$$ and $$\xi \ge \beta$$, i.e. that if $$M_\delta$$ and $$N_\xi$$ are comparable via some $$G$$, then $$M_\delta$$ falls below $$N_\xi$$.(**)

(*) and (**) are the two points where I can't finish this argument. Now my question is that, can the above argument be completed? Or is there some other way to prove that $$\lt_{ip}$$ is well-defined?

Also I would really appreciate any hints or remarks concerning the well-order part.

EDIT I:

The material here can be found in Kanamori's "The Higher Infinite", page $$273$$, $$2$$nd edition.

EDIT II:

It kindly was pointed out to me at MO in the comments by Yair Hayut that the definition of $$\lt_{ip}$$ here is a little bit flawed. I fixed it.

Also it was pointed out that it is reasonable to identify each premouse with it's iterates. In this light $$(1)$$ becomes:

$$(1)^*$$ In the case $$\zeta = \eta$$ we should find some premouse $$\langle B, \in, O\rangle$$ such that both $$M$$ and $$N$$ are iterates of $$B$$. In this case we insure totality.

• @GEdgar "Premouse" is a technical term, and "premice" is plural for it. (Should probably be spelled "pre-mouse" and "pre-mice" to avoid this confusion.) Oct 30, 2019 at 13:00
• @GEdgar: In logic this is called a false premise. Oct 30, 2019 at 14:31
• @AsafKaragila And it's an easy mistake to repeat: it's an iterable false premise. Oct 30, 2019 at 15:04
• For that matter, irritable premice. Feb 3, 2020 at 12:10
• @WillJagy: The correct pun here is "a terrible premice"... (It does work better when considering a terrible mouse, though. But then again, a mouse is just a terrible premouse. So all mice are terrible by definition...) Feb 4, 2020 at 12:50